$\overline{AB}$ = $3\sqrt{13}$ $\overline{BC} = {?}$ $A$ $C$ $B$ $3\sqrt{13}$ $?$ $ \sin( \angle BAC ) = \frac{3\sqrt{13} }{13}, \cos( \angle BAC ) = \frac{2\sqrt{13} }{13}, \tan( \angle BAC ) = \dfrac{3}{2}$
$\overline{AB}$ is the hypotenuse $\overline{BC}$ is opposite to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle BAC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{BC}}{\overline{AB}}= \frac{\overline{BC}}{3\sqrt{13}} $ $ \overline{BC}=3\sqrt{13} \cdot \sin( \angle BAC ) = 3\sqrt{13} \cdot \frac{3\sqrt{13} }{13} = 9$